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Predict the sign of \(\Delta S_{\mathrm{p}}\) for each of the following processes: (a) Molten gold solidifies. (b) Gaseous \(\mathrm{Cl}_{2}\) dissociates in the stratosphere to form gaseous \(\mathrm{Cl}\) atoms, (c) Gaseous \(\mathrm{CO}\) reacts with gaseous \(\mathrm{H}_{2}\) to form liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\). (d) Calcium phosphate precipitates upon mixing \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\). Entropy Changes in Chemical Reactions (Section 19.4)

Short Answer

Expert verified
(a) For the solidification of molten gold, the entropy decreases, so ΔS < 0. (b) For the dissociation of gaseous Cl2 into gaseous Cl atoms, the entropy increases, so ΔS > 0. (c) For the reaction of gaseous CO and H2 to form liquid methanol, the entropy decreases, so ΔS < 0. (d) For the formation of calcium phosphate precipitate, the entropy decreases, so ΔS < 0.

Step by step solution

01

Process (a) - Molten gold solidifies

When molten gold, which is in the liquid state, solidifies into a solid, it experiences a transition from a more disordered state to a more ordered state. This is because particles in the solid state have less freedom of movement and are more structured in comparison with those in the liquid state. Therefore, in this process, the entropy decreases, and we can predict the sign of ΔS to be negative (ΔS < 0).
02

Process (b) - Gaseous Cl2 dissociates into gaseous Cl atoms

In this case, we have gaseous Cl2 molecules splitting into two gaseous Cl atoms. This process leads to an increase in the number of particles in the system, hence increasing the disorder or randomness. Consequently, this process leads to an increase in entropy, and we can predict the sign of ΔS to be positive (ΔS > 0).
03

Process (c) - Gaseous CO reacts with gaseous H2 to form liquid methanol (CH3OH)

In this reaction, two gaseous reactants (CO and H2) combine to form a single liquid product (CH3OH). This process results in a reduction in the number of particles, as well as a transition from the gaseous state to the liquid state, which is more ordered. Both factors contribute to a decrease in entropy. Therefore, we can predict the sign of ΔS for this process to be negative (ΔS < 0).
04

Process (d) - Formation of calcium phosphate precipitate

The formation of a precipitate involves the mixing of two aqueous solutions, Ca(NO3)2 and (NH4)3PO4, to form solid calcium phosphate. This process involves a transition from the more disordered aqueous state to the more ordered solid state. As a result, the entropy of the system decreases, and we can predict the sign of ΔS to be negative (ΔS < 0).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Predicting Entropy Sign
To better understand chemical reactions, it's critical to grasp the concept of entropy and to predict the sign of entropy change, represented by . Entropy is a measure of the disorder or randomness in a system, and during a chemical reaction or physical process, the entropy can increase or decrease. The sign of changes signifies whether the process leads to greater disorder (positive ) or to a reduction in disorder (negative ). Generally, an increase in the number of gas molecules or a move from a more structured phase to a less structured one results in positive entropy change. Conversely, when a system becomes more ordered—such as a gas condensing to a liquid or a reaction producing fewer gaseous molecules—the entropy typically decreases, indicated by a negative sign.

For educational purposes, always consider the state of matter and the number of particles before and after the reaction to predict the sign of effectively. For instance, the solidification of molten gold indicates a transition from a disorderly liquid to a well-arranged solid, predicting a negative entropy change, as seen in Process (a) in the exercise.
Phase Transitions
Phase transitions play a significant role in determining the entropy change during a chemical reaction. The phases of matter—solid, liquid, and gas—have different degrees of order and freedom for the constituent particles. During a phase transition, such as melting, freezing, evaporation, or condensation, the entropy of a system will change. A solid-to-liquid transition, as in melting, increases entropy due to the increased freedom of particles, whereas the reverse, solidification, leads to a decrease in entropy due to reduced randomness, as illustrated by the solidification of molten gold (Process a).

Understanding the nature of phase transitions is paramount to predict the entropy change correctly. Keep in mind that the formation of a gas indicates a higher entropy, while transitions into solids or liquids point to lower entropy, explaining the changes in Process (c) and (d) where gaseous reactants form a liquid or a solid precipitate.
Chemical Reaction Spontaneity
The spontaneity of chemical reactions is often analyzed through the lens of entropy and Gibbs free energy change ( ). A spontaneous reaction is one that proceeds on its own under a given set of conditions. Entropy is one of the key players in this determination, alongside enthalpy ( ). Spontaneous processes tend to increase the overall entropy of the universe, which is the sum of the entropy changes of the system and surroundings. If a reaction results in an increased entropy within the system and the process is exothermic ( < 0), it typically favors spontaneity. Therefore, the sign of the entropy change, as well as the temperature and enthalpy change, must be accounted for when assessing the spontaneity of a reaction. For example, in Process (b), where gaseous dissociates, the positive entropy change supports spontaneity under certain conditions.
Disorder and Randomness in Systems
The concepts of disorder and randomness are at the heart of understanding entropy in chemical processes. Entropy quantifies the degree of uncertainty or the number of possible arrangements the particles in a system can have. A highly ordered system, such as a crystalline solid, has low entropy due to the structured arrangement of its particles. In contrast, gases exhibit high entropy because their particles are spread out and can occupy many positions, leading to a high degree of randomness.

When a reaction or phase change results in a greater spread of particles—such as in the dissociation of into atoms in Process (b)—the system's disorder increases, indicating a rise in entropy. Always consider how the arrangement and movement of particles change during a reaction to assess the effect on the system's disorder. This will not only help you predict the direction of entropy change but also enhance your understanding of the system's behavior under different conditions.

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Most popular questions from this chapter

(a) Does the entropy of the surroundings increase for spontancous processes? (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {gar }}\) ? (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {sury }}=-78 \mathrm{~J} / \mathrm{K}\). What is the entropy change of the system for this process?

Indicate whether cach statement is true or false. (a) The second law of thermodynamics says that entropy is conserved. (b) If the entropy of the system increases during a reversible process, the entropy change of the surroundings must decrease by the same amount. (c) In a certain spontaneous process the system undergoes an entropy change of \(4.2 \mathrm{~J} / \mathrm{K}\); therefore, the entropy change of the surroundings must be \(-4.2 \mathrm{~J} / \mathrm{K}\).

Consider the vaporization of liquid water to steam at a pressure of \(1 \mathrm{~atm}\). (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

|Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix \(\mathrm{C}\). You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4}\) ? (c) At what temperature will an equilibrium mixture of 10 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4}\) ? (d) Rationalize the results from parts (b) and (c) by using Le Chatelier's principle. [Section 15.7]

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{t}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix C, calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{e}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term (-TUS)? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

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